# Thread: Help with Proof of Sets

1. ## Help with Proof of Sets

Prove $(A \cup B) - (A \cap B) = (A-B) \cup (B-A)$

$(A \cup B) - (A \cap B) \subseteq (A-B) \cup (B-A)$
Let $x \in (A \cup B) - (A \cap B)$. Then $x \in A \cup B$ and $x \notin A \cap B$. Then $x \in A$ or $x \in B$ but $x \notin A$ and $x \notin B$. Since $x \in A$ and $x \notin B$ then $x \in (A-B)$. Also since $x \in B$ and $x \notin A$ then, $x \in (B-A)$. Now $x \in (A-B)\cup(B-A)$.

Therefore, $(A \cup B) - (A \cap B) \subseteq (A-B) \cup (B-A)$

Do I have the general idea right? What can I fix? I'm sure I did something wrong. Yes, I also realized I skipped the other part of the proof.

2. Hello, JSB1917!

I used a two-column proof . . .

$\text{Prove: }\;(A \cup B) - (A \cap B) = (A-B) \cup (B-A)$

$\begin{array}{ccccccc}
1. & (A \cup B) - (A \cap B) && 1.& \text{ Given} \\ \\
2. & (A \cup B) \cap (\overline{A \cap B}) && 2. & \text{d{e}f. Subtraction} \\ \\
3. & (A \cup B) \cap (\overline A \cup \overline B) && 3. & \text{DeMorgan's Law} \\ \\
4. & (A \cap \overline A) \cup (A \cap \overline B) \cup (B \cap \overline A) \cup (B \cap \overline B) && 4. & \text{Distributive Prop.} \\ \\
5. & \emptyset \;\;\cup (A \cap \overline B) \cup (B \cap \overline A) \cup \;\;\emptyset && 5. & S \cap S \:=\:\emptyset \\ \\
6. & (A \cap \overline B) \cup (B \cap \overline A) && 6. & S \cup \emptyset \:=\:S \\ \\
7. & (a - B) \cup (B - A) && 7. & \text{d{e}f. Subtraction}
\end{array}$